Given an array of size n, find the majority element. The majority element is the element that appears more than ⌊ n/2 ⌋
times.
You may assume that the array is non-empty and the majority element always exist in the array.
A rather straight forward method is to use hashmap. However, it is a waste of memory space.
This is actually a pretty famous algorithm called: MJRTY - A Fast Majority Vote Algorithm by Robet S. Boyer.
#include <vector> #include <iostream> #include <climits> using namespace std; int majorityElement(vector<int>& nums) { int count = 0; int target = INT_MAX; int i = 0; while(i < nums.size()) { if(count == 0) { target = nums[i]; count++; } else if(nums[i] == target){ count++; } else { count--; } i++; } count = 0; for(i = 0; i < nums.size(); ++i) { if(nums[i] == target) count++; } if(count > nums.size() / 2) return target; else return -1; // remember to verify, it maybe has no majority. For example [1, 1, 2, 2] } int main(void) { vector<int> nums{1, 1, 2}; int target = majorityElement(nums); cout << target << endl; }